If $(f_n)$ is an orthonormal basis and $\lambda_n\to\infty$, is $T(t)x:=\sum_ne^{-λ_nt}\langle x,f_n\rangle f_n$ uniformly continuous?

Question

Let $H$ be a complex Hilbert space, $(f_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$ and $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ be nondecreasing with $\lambda_n\xrightarrow{n\to\infty}\infty$. It is easy to see that $$T(t)x:=\sum_{n\in\mathbb N}e^{-\lambda_nt}\langle x,f_n\rangle_Hf_n\;\;\;\text{for }x\in H\text{ and }t\ge0$$ is a well-defined contractive semigroup on $H$.

Are we able to show that

1. $(T(t))_{t\ge0}$ is uniformly continuous?
2. If $U$ is a complex Hilbert space compactly and densely embedded into $H$ and $$e_n:=\frac1{\sqrt\lambda_n}f_n\;\;\;\text{for }n\in\mathbb N$$ is an orthonormal basis of $U$, does it follow that $$S(t)u:=T(t)u\;\;\;\text{for }u\in U\text{ and }t\ge0$$ is a strongly/uniformly semigroup on $U$ as well?

Noting that $$\sup_{t\ge0}\left\|\sum_{i=1}^ne^{-\lambda_it}\langle x,f_i\rangle_Hf_i-T(t)x\right\|_H^2\le\sum_{i>n}\left|\langle x,f_i\rangle_H\right|^2\xrightarrow{n\to\infty}0\tag1$$ for all $x\in H$, we should be able to conclude that $(T(t))_{t\ge0}$ is at least strongly continuous.

Answer 1

It is not uniformly continuous: $$ \frac1t (T(t)f_i-f_i) = \frac1t (e^{-\lambda_i t}-1)f_i \to -\lambda_i f_i $$ for $t\to0$. The operator $f_i \mapsto \lambda_i f_i$ is unbounded, so $T$ cannot be uniformly continuous. Or put differently, the convergence $\frac1t (T(t)f_i-f_i) \to -\lambda_i f_i$ is not uniform with respect to $i$, and $\frac1t\|T(t)-I\|$ does not exist.